已知函数f(x)=4sin^x+2cos(2x-π/3),

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已知函数f(x)=4sin^x+2cos(2x-π/3),
（1）令f(x1)=3,求x1的值构成的集合
（2）若x∈[0,π/2],f(x)=5/2,求sin2x的值

sin^2 x = 1/2 (1-cos2x)
so f(x) = 2(1-cos2x) + 2 cos(2x-pi/3) = 2 + 2(cos(2x-pi/3)-cos2x))
= 2 + 2 (cos2xcospi/3 + sin2x sinpi/3 - cos2x) = 2+2(1/2 cos2x + sqrt(3)/2 sin2x) = 2+2sin(2x+pi/6)
(1) f(x1) = 3,sin(2x+pi/6) = 1/2,2x + pi/6 = 2npi + pi/6 or 2npi + 5pi/6
2x = 2npi or 2x = 2npi + 2pi/3
x = npi or x = npi + pi/3
(2) f(x) = 5/2,sin(2x+pi/6) = 1/4
sin2x sqrt(3)/2 + cos2x *1/2 = 1/4
sin2x sqrt(3) + cos2x = 1/2
sin2x sqrt(3) - 1/2 = -cos2x
let sin2x = y and square both sides
(sqrt(3)y - 1/2)^2 = 1-y^2
3y^2 - sqrt(3)y + 1/4 = 1 - y^2
4y^2 - sqrt(3)y - 3/4 = 0
y = [sqrt(3) +/- sqrt(3 + 12)] / 8 = [sqrt(3) +/- sqrt(15)] / 8
since x is in [0,pi/2],sin2x > 0
so sin2x = [sqrt(3)+sqrt(15) ]/ 8

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